Stability 1 – Chapter 5 Effect of Density Solutions
- A box-shaped vessel 18 x5 x2 m floats in DW of RD1.000 at a draft 1.4m. Calculate her percentage reserve buoyancy when she enters the SW .
Solution:
Volume of box shape vessel = (L x B x H )
= (18m x 5m x 2m)
RD of DW = 1.000 & Depth = 1.4m
Displacement at DW of RD 1.000 (W)
= (u/w volume)x (density)
=( Lx B x 1.4) x(1)
Displacement at SW (W1) = (u/w volume) x (density)
=(L x B x Draft) x (1.025)
Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as MASS.
W = W1
(L x B x 1.4) x (1) = (L x B x Draft) x (1.025)
Hence, Draft = (1.4/1.025)
= 1.365m
Total volume of the ship = (L x B x H)
= (18 x 5 x 2)
= 180m3
Again, U/w volume of SW = (L x B x draft)
=(18 x 5 x 1.365)
= 122.85m3
Above water volume = (180 -122.85)
= 57.15m3
Hence, RB % = (Above water volume/ Total volume) x 100
= (57.15/ 180) x 100
= 31.75%.
- The hydrostatic particular of a ship indicate that her displacement in SW at a draft of 5m is 3000t. Find her displacement when floating at 5m draft in water of RD 1.018.
Solution:
Displacement(W) = 3000t,
Draft= 5m & RD = 1.018
W = (u/w volume) x (density)
= (L x B x draft) x(1.025)
Draft = W/ (L x B x 1.025)
Let W1 be the displacement at RD = 1.018
W1 = ( L x B x d) x (1.018 )
d1 = W1/( Lx B x 1.018)
But according to question ,
Draft = d1
W/ (L x B x1.025 )= W1 /(L x B x 1.018)
3000 / 1.025 = W1 / 1.018
W1 = 3000 x (1.018 / 1.025)
= 2979.51 t
- A vessel displaces 4500 t of FW at a certain draft . Find her displacement at the same draft in water of RD 1.020 .
Solution:
Displacement (W) = 4500t,
Displacement = (u/w volume ) x (density)
= (L x B x draft) x 1.025
Draft = W / (L x B x 1.025) m
Let W1 be the displacement at RD 1.020
So W1 =( L x B x d )x (1.020)
d = (W1 / (Lx B x 1.020)
according to question, Draft = d
W /( L x B x 1.025) = W1 /( L x B x 1.020)
W1 = (4500 x 1.020) / 1.025
= 4478.04t.
- A ship 100m long and 20m wide, block coefficient 0.8m, floats in SW at a mean draft of 8.0 m. Calculate the difference in displacement when floating at the same draft in FW.
Solution:
Area of Waterplane = (L X B)
= (100m x 20m)
Cb = 0.8, Depth = 8m
We know that :
Displacement = (u/w volume ) x (density)
= ( Lx B x draft ) x (0.8)
= (100 x 20 x 8) x(0.8)
= 12800m3
Displacement (W) = (u/w volume)x (density)
= (12800 x 1.025)
= 13120t ,
Draft = W / ( L x B x Cb x 1.025)
Let W1 be the displacement in FW
= (u/w volume) x (density)
=(L x B x d1 x Cb ) x ( density)
d1 = (W1 /( L x B x 1 )x (Cb)
According to question, Draft = d1
W /( L x B x Cb x 1.025) = ( W1 /( L x B x Cb)
W1 = ( 13120 x 1) /( 1.025)
= 12800t
So, Difference is displacement = (13120 – 12800)
= 320 t .
- A vessel displaces 14500 tonnes, if floating in SW up to her winter load- line. If she is in a dock of RD 1.010 , with her winter load- line on the surface water , find how much cargo she can load, so that she would floats at her winter load-line in SW.
Solution:
Displacement (W )= 14500t
RD = 1.025
Displacement (W) =( u/w volume ) x (density)
= (L x B x draft) x (1.025)
Draft = W /( L x B x 1.025)
Let W1 be the displacement at RD of 1.010
So W1 = (u/w volume )x( density)
= (L x B x d1 ) x (1.010)
Hence, d1 = W1 / (L x B x 1.010)
According to question, Draft = d1
So, W / ( L x B x 1.025) = W1 / (L x B x 1.010)
14500 / (L x B x 1.025) = W1 / (L x B x 1.010)
W1 = (14500 x 1.010) / 1.025
= 14287.8 t
Hence, Cargo to load = (14500 – 14287.8)
= 212.2t.
- A vessel of 12000 t displacement arrives at the mouth of a river , drawing 10.0 m in SW. how much cargo must she discharge so that her draft in an up river port of RD 1.012 would be 10m .
Solution:
Displacement (W)= 12000t ,
Depth = 10m & RD = 1.025
Let W1 be the displacement at RD 1.012
Now, W =( L x B x draft ) x( 1.025)
W1 = (L x B x d1) x (1.012)
According to question ,
Draft = d1
W/( L x B x 1.025) = W1 / (L x B x 1.012)
W/ (L x B x 1.025) = W1/ ( L x B x 1.025)
W1 = (12000 x 1.012) / 1.025
W1 = 11847.8 t
Hence, Cargo she has to discharge = (12000 – 11847.8) t
= 152.2 t
- A vessel floating in DW of RD 1.005 has the upper edge of her summer loadline in the water line to starboard and 50mm above the waterline to port . If her FWA is 180mm and TPC is 24 , find the amount of cargo which the vessel can load to bring her to her permissible draft .
Solution:
RD = 1.0
Starboard side = 0cm above by below summer draft
Port side = 50mm (5cm above the water line to summer draft)
Mean draft = (0+5 ) / 2
= 5/ 2
= 2.5cm above the water line
FWA = 180mm
= 18cm,
TPC = 24
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.005 ) x 18 /0 .025
= (0.2 x 18) /0 .025
= 14.4cm
Total sinkage = (14.4 + 2.5) cm
= 16.9cm
TPC is SW = 24
TPC = (A / 100) x 1.025
= (24 x100)/ 1.025
A = 23.41 t/cm
Again ,
TPC in RD 1.005 = ( A / 100) x 1.005
= (23.41 x 1.005)
= 23.53t/cm
Cargo to load = sinkage x TPC
= 23.53 x 16.9
= 397.657 t
- A vessel is floating at 7.8m draft in DW of RD 1.010 . TPC is 18 and FWA is 250mm. the maximum permissible draft .
Solution:
Present draft = 7.8m,
RD of DW = 1.010 & TPC = 18 t/cm
FWA = 250mm = 25cm
Maximum permissible draft = 8.0m
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.010 – 1.020 ) x FWA /0 .025
=(0.015 x 25) /0 .025
= 15cm
When vessel enters SW, the draft will rise by 15 cm,
Hence ,actual draft in SW = (7.80 -0. 15)
=7.65m
Sinkage available = (8.00 – 7.65)
=0.35m
TPC = 18 ( given)
We know that :
TPC = (A/100) x ( density)
18 = (A/100) x 1.025
So, A = (18 x 1.025)/ 100
= 17.56 m2
Now , TPC at RD 1.010 = A / 100 x 1.010
= 17.56 x 1.010
= 17.73 t/cm.
DWT available =( sinkage x TPC)
= (35 x 17.73)
= 620.5t.
- A vessel’s statutory freeboard is 2.0m. She is loading in DW of RDD 1.015 and her freeboard is 2.1m . TPC =24. FWA = 200mm. find the DWT available.
Solution:
Statutory freeboard is = 2.0m
DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.015 – 1.025 ) x 20 /0.025
= ( 0.010 x 20) /0 .025
= 8cm
Hence , actual freeboard available =( 2.1 + 0.08)m
= 2.18m
Sinkage available = ( 2.18 – 2.00)
= 0.18m
= 18cm
TPC = 24 ( Given)
24 = (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A = (24 / 1.025)
= 23.414 m2
Now, TPC for RD 1.015 = (A/ 100) x(1.015)
= (23.414 x 1.015)
= 23.76 t/cm
DWT available = (sinkage x TPC)
= (18 x 23.76 )
= 427.68t.